Let A(3, 7) and B(6, 5) are two points. `C:x^(2)+y^(2)-4x-6y-3=0` is a circle.
Q. If O is the origin and P is the center of C, then absolute value of difference of the squares of the lengths of the tangents from A and B to the circle C is equal to :

Options

A. `(AB)^(2)`
B. `(OP)^(2)`
C. `|(AP)^(2)-(BP)^(2)|`
D. `(AP)^(2)+(BP)^(2)`

Answer

To solve the problem, we need to find the absolute value of the difference of the squares of the lengths of the tangents from points A(3, 7) and B(6, 5) to the circle defined by the equation \( C: x^2 + y^2 - 4x - 6y - 3 = 0 \).

### Step 1: Find the center and radius of the circle
The equation of the circle can be rewritten in standard form by completing the square.

1. Rearranging the equation:
\[ x^2 - 4x + y^2 - 6y = 3 \]

2. Completing the square for \(x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]

3. Completing the square for \(y\):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

4. Substitute back into the equation:
\[ (x - 2)^2 - 4 + (y - 3)^2 - 9 = 3 \]
\[ (x - 2)^2 + (y - 3)^2 = 16 \]

From this, we can see that the center \(P\) of the circle is at \( (2, 3) \) and the radius \(r\) is \(4\) (since \(r^2 = 16\)).

### Step 2: Calculate the lengths of the tangents from points A and B
The length of the tangent from a point \((x_1, y_1)\) to a circle with center \((h, k)\) and radius \(r\) is given by the formula:
\[ L = \sqrt{(x_1 - h)^2 + (y_1 - k)^2 - r^2} \]

#### Length of tangent from point A(3, 7):
1. Substitute \(A(3, 7)\) into the formula:
\[ L_A = \sqrt{(3 - 2)^2 + (7 - 3)^2 - 4^2} \]
\[ = \sqrt{1^2 + 4^2 - 16} \]
\[ = \sqrt{1 + 16 - 16} = \sqrt{1} = 1 \]

#### Length of tangent from point B(6, 5):
2. Substitute \(B(6, 5)\) into the formula:
\[ L_B = \sqrt{(6 - 2)^2 + (5 - 3)^2 - 4^2} \]
\[ = \sqrt{4^2 + 2^2 - 16} \]
\[ = \sqrt{16 + 4 - 16} = \sqrt{4} = 2 \]

### Step 3: Calculate the squares of the lengths of the tangents
1. Square the lengths:
\[ L_A^2 = 1^2 = 1 \]
\[ L_B^2 = 2^2 = 4 \]

### Step 4: Find the absolute value of the difference of the squares
1. Calculate the absolute difference:
\[ |L_A^2 - L_B^2| = |1 - 4| = | -3 | = 3 \]

### Final Answer
The absolute value of the difference of the squares of the lengths of the tangents from A and B to the circle C is \(3\).

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